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        riddles >> easy >> A 2006 Puzzle
        
(Message started by: K Sengupta on Jun 25^th, 2006, 7:54pm)

Title: A 2006 Puzzle
Post by K Sengupta on Jun 25^th, 2006, 7:54pm

Determine the total number of positive integer solutions of this equation:

2/x + 3/y = 1/2006.

Title: Re: A 2006 Puzzle
Post by jollytall on Jun 26^th, 2006, 4:45am

Well, not the complete solution, simply some equation manipulation gave the way to all solutitions.

[hide]2/x + 3/y = 1/2006

2*17*59*(2y+3x) = x*y

X and y shall have 2, 17, and 59 as dividers. There are seven options:
X=2*17*59*x' => y has no restrictions
x=2*17*x', y=59*y'
x=17*59*x', y=2*y'
x=59*2*x', y=17*y'
x=2*x', y=17*59*y'
x=17*x', y=59*2*y'
x=59*x', y=2*17*y'
x is free, y=2*17*59*y'

If we try them all:

First equation:
2*y+3*2*17*59*x'=x'*y
y=(3*2006*x')/(x'-2) | X'>2
There are a number of solutions, where x'-2 is dividing 2*3*17*59 (x' does not count, since there are only two numbers, 3 and 4, where x'-2 divides x' too, but those are covered anyway:
x'=3, x=3*2006, y=9*2006
x'=4, x=4*2006, y=6*2006
x'=5, x=5*2006, y=5*2006
etc.

Second equation:
2*59*y'+3*2*17*x'=x'*y'
y'=(3*2*17*x')/(x'-2*59)
Again there are many solutions x'>2*59:
x'=119, x=4046, y'=12138, y=716142
etc.

For all equations we can do it.

The last one:
y'=3*x/(x-2*2006)
The first few solutions:
x=4013, y'=12039, y=24150234
x=4014, y'=6021, y=12078126
x=4015, y'=4015, y=8054090
x=4016, y'=3012, y=6042072
etc.[/hide]
There might be a more elegant solution, but I still don't have it.

Title: Re: A 2006 Puzzle
Post by Icarus on Jun 26^th, 2006, 3:32pm

Try the substitutions: r = x - 4012, s = y - 6018.

Title: Re: A 2006 Puzzle
Post by K Sengupta on Jun 26^th, 2006, 11:08pm

On Today at 3:32pm , Icarus wrote:

Quote:

Try the substitutions: r = x - 4012, s = y - 6018.

I would like to thank Icarus for his invaluable guidance in the matter.

Substituting, r = x - 4012, s = y - 6018., we obtain:

r*s = 6*2006² = 3*2³*17²*59²

Accordingly, the required number of solutions
= The total number of factors of 6*2006²
= 2*4*3*3
=72.

This corresponds to the case when x> 4012 and y>6018, and does not take into account the pairs (x,y) satisfying 1<=x<= 4012 and 1<=y<=6018.

But when 1<=x<=4012, we obtain 2/x>= 2/4012 = 1/2006, which is a contradiction, since no positive value of y is then feasible in conformity with tenets corresponding to the problem under reference.

Similarly, when 1<=y<=6018, we obtain 3/y>= 3/6018 = 1/2006, which is a contradiction, since no positive value of x is then possible for obvious reasons.

Combining all the available cases, we must conclude that the required number of positive integer solutions to the given problem = 72.

Title: Re: A 2006 Puzzle
Post by Barukh on Jun 26^th, 2006, 11:29pm

on 06/26/06 at 23:08:08, K Sengupta wrote:

However, it would be observed that this number only corresponds to the number of solutions whenever x> 4012 and y>6018, and does not take into account the pairs (x,y) satisfying 1<=x<= 4012 and 1<=y<=6018.

If x <= 4012, then 2/x >= 1/2006, and so no positive y yields the required equality.

Title: Re: A 2006 Puzzle
Post by K Sengupta on Jun 27^th, 2006, 12:48am

On, Jun 26th, 2006, 11:29pm Barukh wrote:

Quote:

If x <= 4012, then 2/x >= 1/2006, and so no positive y yields the required equality.

I wish to thank Barukh for obviating one of my doubts regarding solution to the problem under reference.

In light of the foregoing, I confirm having suitably amended my previous post (Jun 26th, 2006, 11:08pm) to reflect the precise solution to the given problem.